\(\frac{d}{dx}\ arcsin(x)\)

\begin{align}Let \, y & = arcsin(x)\\ then \quad x & = sin(y) \quad \enclose{circle}{\color{black}{1}}\\ And \quad \frac{dx}{dy} & = cos(y) \\ Inverting \quad \frac{dy}{dx} & = \frac{1}{cos(y)} \\ & = \frac{1}{\sqrt{1-sin^2(y)}} \\ Substituting \, from \, \enclose{circle}{\color{black}{1}}\\ & = \frac{1}{\sqrt{1-x^2}} \end{align}

\( \frac{d}{dx}\ arccos(x)\)

\begin{align}Let \, y & = arccos(x)\\ then \quad x & = cos(y) \quad \enclose{circle}{\color{black}{1}}\\ And \quad \frac{dx}{dy} & = -sin(y) \\ Inverting \quad \frac{dy}{dx} & = -\frac{1}{sin(y)} \\ & = -\frac{1}{\sqrt{1-cos^2(y)}} \\ Substituting \, from \, \enclose{circle}{\color{black}{1}}\\ & = -\frac{1}{\sqrt{1-x^2}} \end{align}

\(\frac{d}{dx}\ arctan(x) \)

\begin{align}Let \, y & = arctan(x)\\ then \quad x & = tan(y) \quad \enclose{circle}{\color{black}{1}}\\ And \quad \frac{dx}{dy} & = sec^2(y) \\ Inverting \quad \frac{dy}{dx} & = \frac{1}{sec^2(y)} \\ & = \frac{1}{1+tan^2(y)} \\ Substituting \, from \, \enclose{circle}{\color{black}{1}}\\ & = \frac{1}{1+x^2} \end{align}